We can model these phenomena by the equation dy/dt = ky where:
y(t) is the value of the quantity at time t,
k is a constant,
dy/dt is a rate of change of y with respect to t.
If k > 0, then dy/dt > 0 is the GROWTH rate.
If k < 0, then dy/dt < 0 is the DECAY rate.
The above equation is a differential equation because it involves an unknown function y and its derivative dy/dt. The solutions of the differential equation dy/dt = ky are y(t) = y(0).exp(kt) where y(0) is the initial value.
In exponential growth (k > 0), the rate of change increases over time - the rate of the growth becomes faster as time passes. In exponential decay (k < 0), the rate of change decreases over time - the rate of the decay becomes slower as time passes.
Depending on your major or area of interest, please choose ONE example in Section 3.8 of the textbook and answer the following questions:
1) Specify the number of the example you chose (example 1, example 2, etc).
2) In this example, please analyze how the above model is used.
3) Is your chosen example exponential growth or decay?
If you do not have the textbook, you can look at the examples of Section 3.8 on the class website.
The deadline to submit your comments is Thursday, 11/15. Late submission will be graded zero.
PLEASE WRITE YOUR NAME AT THE BOTTOM OF YOUR COMMENTS FOR GRADING. IF YOU HAPPEN TO FORGET YOUR ACCOUNT ID OR PSW, GO TO GOOGLE TO CREATE A NEW ONE. DO NOT SEND YOUR COMMENTS TO ME BY EMAIL!
29 comments:
1) I chose Example 1.
2) This specific example deals with population growth rate of a specific time period.In the example they began the experiment by staring with the year 1950 which at that time the population was 2,560. They labeled (P) as population (in millions) and (t) for time. For the beginning year 1950 they set the time (t) to 0.Then to find the population growth they used (t) and represented that number to be 10, to find the population growth rate 10 years after 1950. They then used the derivative dP/dt = kP. They then plugged in the correct values for each variable.For example, P(10) = 2,560e^10k = 3040. Then they plugged in 10 for t and took the natural log of the equation to find the calculated value for k. Which in this particular equation was k~0.017185. So the growth rate was calculated to be approximately 1.7% per year. So the final model for this experimant is P(t)= 2560e^0.017185t. So to better answer their question about the population growth rate in the future they calculated the population growth rate for 43 years after 1950 which is 1993. They then used the model P(43)=2560e^0.017185(43). The population was claculated to be approximately 5360 million.They then calculated the population for 2020 and the number was even higher. If the population grows higher this can become a major problem for our economy because the worlds resources are going to diminish quickly. This is a good way to calculate the population growth rate.
3) In this particular experiment this is an example of exponential growth. If you were to graph the data that was calculated in the experiment you will see the population growing exponentially. K>0 and dy/dt >0 , the rate of growth is faster as times passes.
~Casey Mauldin ~~
1) Example 3
2) This example involves Newton's Law of Cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. If T(t) is the temperature of the object at time t and Ts is the temperature of the surroundings, then we have this equation:
dT/dt = k(T-Ts)
Since Ts is constant, it can change to y'(t)=T'(t) and the equation becomes:
dy/dt = ky
The problems states that a bottle of soda at 72 degrees is in a refrigerator at 44 degrees and after half an hour, it cooled to 61 degrees. What is the temp after another half hour?
We can set up dT/dt = k(T-44)
If we let y = T-44, then y(0) = T(0) - ff = 72 - 44 = 28, so:
dy/dt = ky y(0)=28
Using theorem 2 in the textbook, that becomes:
y(t) = y(0)e^kt = 28^kt
We are given T(30)=61, so y(30)=61-44=17 so:
28e^30k = 17
e^30k = 17/28
k = ln(17/28) / 30
k = 0.01663
Now that we have k, we have to plug it back into the equation to find the temp after another half an hour:
T(t) = 44 + 28e^-.01663t
T(60) = 44 + 28e^-.01663(60)
= 54.3
So after another half hour, or a total of one hour, the soda cooled to 54 degrees.
3) This example is exponential decay because the temperature is decreasing at the soda cools.
Rachel Discavage
1) Example 1 Population Growth
2) The population growth problem used in example one is an illustration of exponential growth. In order to calculate what the relative growth rate of the population would be in the second half of the twentieth century a model is set up using given information of past population size. It is assumed that the growth rate is proportional to the population size. The given information is that in 1950 the population size was 2560 million and ten years later in 1960 it was 3040 million. Time(t) is measured in years beginning at 1950. So t=0 in 1950. Population,P(t), is measured in millions of people. This gives P(0)=2560 and P(10)=3040. Using the theorem,
P(t)=Poe^kt, and putting in the information to find k this produces,
P(t)=P(0)e^kt=2560e^kt
P(10)=2560e^10k=3040
Take the natural log of both sides
k=1/10 ln(3040/2560)=0.017185
so the relative growth rate is about 1.7% each year. Now the equation can be written as
P(t)=2560e^0.017185t
In order to calculate the population in 1993 we subtract 1950 and 1993 to get 43 years. Now we can plug it into the equation.
P(43)=2560e^0.017185(43)=5360 million. The same process is done to calculate the population in 2020.
P(70)=2560e^0.017185(70)=8524 million.
3)Graphing the results would give visual conformation that this is exponential growth.
Karen Sanders
Example 3 - Newton's Law of Cooling
2) In this example, the soda pop is cooling, so it's exponential decay. The equation y(t)=y(0)e^kt is still used. y(t) represents the temperature at the given time, and y(0) is the temperature difference between the object's initial temperature and the surrounding temperature.
The example that I chose is an example of exponential decay when the temperature of an object is cooling, but it is exponential growth when an object is warming.
Jordan Montgomery
1. Example 3
2. In this example the formula used was dT/dt = k(T-Ts). Ts is a constant at 44 F which allows the equation to become dy/dt =ky. From here you can find that y(0) = 28. Then you can plug this into the equation y(t) = y(0)e^kt. Using T(30) = 61 you can find that k = -0.01663 by computing 28e^30k = 17. Then, you can plug k into the original equation T(t)= 44+ 28e^k. Plug in 60 for t to give you another half hour and you get
T(60) = 44 + 28e^-0.01663 =54.3.
3. This example is one of exponential decay because the temperature is decreasing and the temperature decreases at a slower rate as time passes.
william wigglesworth
Example 4:
In this example we can see the exponential growth of our object(in this case a bank account) because as our interest gets compounded(at whatever rate and at whatever interval) you have a steady increase but not at the exact same levels but rather at a steadily increasing rate.
As the graph for exponential growth shows as the time interval(deemed on the X axis) increases, the interval jumps or rate of change increases(as shown on the Y axis)
1) Example 1.
2) Time is reffered to as t. t = 0 at 1950 and t = 10 at 1960. P(t) is the population in millions at t. Here we would use P(t)e^kt. We then insert all our givens to find our rate and set it equal to 3040. After solving this we now have k and with k we have a complete equation with t as a variable. To find for 1993 you insert 1993 -1950 = 43 years and that is our t. As t, which is on the x axis, increases so does the population.
3) It is exponential growth.
Jason Silvestre
a)I chose example 3
b)This is an example of Newton's Law of Cooling, which states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large.In this example they use a bottle of soda at room temperature(72 deg F) and find different temperatures of the same soda after placed in the refridgerator for a certain time. Let T(t) be the soda after t minutes.
Ts=44 deg F, so dT/dt=k(T-44).
If y=t-44, then y(0) would be 28.
y(t)=y(0)e^kt=28e^kt.
Given T(30)=62,
y(30)=61-44=17 and 28e^kt=17 and e^30k=17/28. We take logarithms k=ln(17/28)/30=-.01663. Thus T(60)=44+28e^-.01663(60)=54.3. So after another half hour the pop has coole to about 54 deg F.
For T(t)=50, t=ln(6/28)/-.01663=92.6. Which means that the pop cools to about 50 deg F after 1 hour and 33 minutes.
c)This example shows exponential decay because the temperature is decreasing with time.
Marcus Williams
1.) Example 1 – Population Growth
2.) This problem which illustrates population growth is an example of exponential growth. Given the population size, an equation is set up to calculate the relative growth rate of the second half of the twentieth century. The information is as follows:
1950 – population: 2560 million
1960 – population: 3040 million
Time is measured beginning in 1950 at t=0. Population can be calculated by setting up P(t) for each. This would give us P(0)=2560 and P(10)=3040. Next we would insert our given information into the equation P(t)e^kt and set it equal to 3040. We find that the population growth is about 1.7% a year which we can then insert into our equation as k and solve for t. In order to find the population growth in 1993 all we have to do is subtract 1950-1993 to get 43 years and plug that in for (t). The answer would be 8524 million.
3.) This is an example of exponential growth. This can be confirmed by graphing the equation.
1) Example 2
2) Example 2 deals with the half-life of the radioactive element radium-226. As time passes the element decays and its mass decreases. In this case the half-life is 1590 years. The example wants to determine the mass after t years. To do this we use the given mass (100mg) and half-life to set up the equation m(t)=m(o)e^kt where m(o) is the initial mass. Useing t = 1590years we know that the remaining mass will be 50mg. Setting the euation equal to 50 gives k=-ln(2)/1590. So m(t)=100e^(-ln(2)t/1590). With this model we can find the mass that will remain after any number of years or we can find the number of years it will take to be left with a certain mass.
3) Half-life is an example of exponiential decay because after a certain number of years the mass of the element is cut in half.
Peter Harris
1) Example 1 Population Growth
2) In this example, a growth model is used to show a pattern in population growth in the United States and predict an approximation of the population for some time in more recent history, 1993, and for some future time, 2020.
To make our model, we are given 2 pieces of information: the population in 1950 and the population in 1960. To relate the 2 of them as a function of time, we will let 1950 be represented by t=0. We know that P(0)= 2560 million and P(10)= 3040 million. We are assuming that their relationship will follow the theorem that dP/dt= kP(t) where k>0 because our population is growing.
P(t) is thus P(t)=[P(0)]^(kt), so
P(t)=2560e^kt. Since we know the population in the year 1960, 10 years after 1950, we have:
P(10)= 3040= 2560e^10k
e^10k=3040/2560
10k= ln(3040/2560)
k= (1/10)ln(3040/2560)
k= 0.017185
Now our model is P(t)= 2560e^(0.017185)t
For 1993, 1993-1950= 43
P(43)= 2560e^(0.017185)(43)
= 5360 million
For 2020, 2020-1950= 70
P(70)= 2560e^(0.017185)(70)
= 8524 million
By looking at the graph provided in the example (with our function P(t) and blue dots representing the actual population as recorded every 10 years), we learn that our estimate for the population in 1993 was quite accurate, but only time will tell if our prediction for the year 2020 will be accurate.
3) My chosen is example is exponential growth.
-Katie Moore
1. I choose example 2.
2. The decay rate of any substance after time (t) is m(t)=Moe^kt. Where Mo is the inital mass, k is a negative constant, and t is time. A substance with an inital mass (Mo) of 100 mg and its half life is 1590 years, to find its mass remaining after t years we would then plug these constants into the formula.
m(t)= (100)e^k(1590)
So, 1/2 of 100 is 50 since we are finding the half life. The new formula is: m(t)= e^k(1590)=1/2. If we take the natural log of both sides, we can get rid of the e and the formula would be simplified to 1590k=ln1/2 of 1n-2.
So we now have k= -ln2/1590. We now have made the formula :
m(t)=100e^-ln2(t)/1590. All we have to do is plug in 1000 for t to find th remaining mass of this substance after 1000 years.
m(1000)=100e^-ln2(1000)/1590. After 1000 years, the mass of this substance is 65mg.
3. The topic Radioactive decay, states that this example will be exponential decay, since it relates to half lives of certain substances.
**Stephanie Wallace
1. I chose example 1
2. In this model we took a 10 year period to see the rate of growth, and came up with P(t)=2560e^0.017185t because P(0)e^kt=2560e^kt, and we calculate that k is 0.017185. We took that number and took it to calculate the population in the year 1993 and 2020. So we take that and input for t to come up with the estimated population in t years.
3. It is exponential growth
Trevor Hubbard
1) example 1
2) This example dealt with the growth of population, and showed how this growth of population was simultaneously an example of exponential growth. Given the population size, of both years we can calculate the rate at which the population is increasing. As seen in the example in 1950 2560 million is the population and in 1960 the population rises to 3040 million. Time at the beginning of our period is at 0, because we have not gone anywhere yet, but it eventually will change due to the amount of time. By making time a function of population we can say that P(o)= 2560 and P(10)= 3040. Using these equations, we can use the given information P(t)e^kt and set it equal to our desired year of 3040. By using our equations for and K and solving for T we can see that there is an exponential growth of 1.7%. Like wise we can find the population of 1993 but subtracting the years and plugging that into our equation for time (t). This would give us a population of 8523 million.
3) This can also be verified either by making a table or drawing a graph. The positive slop and rising would show the increase in our population.
1) Example 4.
2) This example deals with continuously compounded interest on something. In this example there is an investment of $1000 and a 6% interest rate. So to calculate how much the investment is worth after a year, find their product, = $1060. To calculate after two years, times it by 1.06 again = $1123.60. We find these examples give off the formula:
Ao(1+r)^t (r = interest rate)
However, you can further calculate shorter times by rewriting the formula:
Ao(1 + (r/n))^(nt)
(n being an integer like .5 for annual or 365 for calculating daily compound interest).
Now if we let n go towards infinity we develop the equation:
Ao[lim (1 + (1/n)^(nt)]^(nt)
But the limit in this equation is equal to e, so it can be rewritten:
A(t) = Aoe^(rt)
Thus making it a compound equation for calculating interest.
This is relative to the graph of Exponential Growth because as the x value gets higher (t), or as the interest is compounded more often, the larger the investment becomes, at a steady pace.
3) Growth
-Jon Kight
1) Example 1
2)This example shows the pattern of how the population of the United States grows. Populations are given from 1950 and 1960, and using the growth model, the population in 1993 and 2020 can be approximated. In the example, 1950 will represent time 0, therefore 1960 will be time 10.
P(0) = 2560 million, and P(10) = 3040 million, and the growth will follow the equation dP/dt = kP(t).
p(t) = p(0)e^(kt), so P(10) = 3040 = 2560e^(10*k).
From this equation we can solve for k = (1/10)ln(3040/2560), about .017185.
Our growth model is P(t) = 2560e^.017185t.
From here we can find the population at 1993 and 2020, by subtracting 1950, getting 43 and 70. Plugging into the growth model, the population in 1993 was 5360 million, and in 2020 it will be 8524 million.
3) From looking at the graph of the model, it shows that the United States population growth is exponential.
Barbra Giourgas
1. Example 2
2. The model shows us how a given mass of radium-226 decays over time. we are given the equation m(t)= m(0)e^kt, since y(0)=100. We then enter the information we have into the formula which gives us: 100e^1590= 50, since we are given that the half-life for y(1590) is 1/2 of 100...
Now we have enough information to solve for K, which equals -ln2/1590.
We now can compute the mass after any number of years, and also the time for that mass (100mg) to be a certain mass. The mass after 1000 years would be m(1000)= 100e^(ln2*1000/1590) which equals about 65 mg.
For the mass of 100mg to be 30, 100e^-(ln*t/15900=30mg sovling for t we get 2762 years.
3. The experiment i chose was one of exponential decay.
Matthew Lombard
1.) Example 1
2.)Example 1 is about the world population growth starting in 1950.
P(t)=P(0)e^kt=2560e^kt
p(10)=2560^10k=3040
k=1/10 ln3040/2560 = .017185
This show a growth rate of 1.7%.
The problem then asks you to estimate the population in 1993 then in 2020. So we can now figure this out.
P(43)=2560e^0.017185(43)=5360million
P(70)=2560e^.017185(70)=8524million
3.) This shows an example of exponential growth.
Matthew Copello
1) I chose Example 1
2)In example one, the graph is used to show population growth/decline over a set amount of years. The time starts recording in 1950 (t=0) and goes until 1960 (t=10). These growths can be used to approximate a the average growth in the future. The population is measured in the millions, p(0)= 2560 and p(10)= 3040. We assume that dP/dt = kP.
p(t) = p(0)e^(kt)
so P(10) = 3040 = 2560e^(10*k)
k=1/10 ln 3040/2560 which is about .017185.
This means the relative growth rate is around 1.7% a year. So to estimate a year of 43 we use the equation p(43)= 2560 e^.017185(43). This will be our assumed population growth.
3. The population grows with time. So this would be called exponential growth and not decay. The graph is up (positive).
Michelle Sokoll
1. I chose example 2
2.Following the theorem for the rate of decay, m(t) = m(0)e^kt where m(0) is the initial mass, k is a negative constant, and t is time; and using the given information to plug into this theorem we get m(t)= 100e^kt where t = 1590 and since we are dealing with half life of the substance our resultant equation that we will be solving for k is 100e^1590k=50. This gives us our k which we can then finally use to find m(t).
3. This deal with half life so this example is demonstrating exponential decay not growth.
-kelly gustafson
I chose example 1.
They compared the two populations from different time periods through a division problem with a constant and the constant e. These type problems are a regular formula P(t) = pe^(kt) and when k is positive, it represents growth. When the k is negative in the exponent, the equation says decay of population.
Jim Prescott
My chosen example was exponential growth because the k came out to be a positive number.
1. Example 1: Growth of World Population
2. In this example, the basic formula P(t)=Pe^(kt) is used to describe the growth of the human population. k is the relative growth rate, t is the number of years, and P is the beginning population. This prediction model can be used to estimate the population at any point past a certain year.
3. In this specific example, because k is positive, the graph of the function will be concave up, meaning that it is an exponential growth model.
Levon Hoomes
I did example number 4. This example is about continuously compounded interest. If the interest is compounded n times a year then it is a rate r/n and there are n*t compounding periods in t years. The value of the investment is A0 (1+r/n0^n*t. Then if we let n approach infinity in the limit A0(1+r/n)^nt = the limit as n approaches infinity A0((1+r/n)^n/t)^rt which equals A0 (limit as n approaches infinity ((1+r/n)^n/r)^rt) which then equals A0 (Limit as m approaches infinity (1+1/m)^m)^rt , where m=n/r. However the limit is equal to e. A(t) = A0ert. Then if we differentiate this we get dA/dt= rA0e^rt=rA(t). This states that, with continuous compounding of interest, the rate of increase of an investment
is proportional to its size. If we returning to the example of $1000 invested for 3 years at 6% interest, we see that, with continuous compounding of interest,
the value of the investment will be. A(3)= 1000e^(.06)3 which equals $1197.22. This is an example of exponential growth.
-Jasper W. Yonker
1. I chose Example 1, for population growth.
2. In this example, the pattern of the population growth in the United States from 1950 to 1960 is shown. t=0 is 1950 and t=10 is 1960. we use the equation dP/dt=kp(t). From the equation p(t)=p(0)e^(kt) we find that the populations in 1950 is 2560 million and in 1960 it is 3040 million. From these equations and numbers it is possible to estimate future US populations. The example wants us to estimate the populations in 1993 and 2020 by using the equation. After plugging into the equation it is possible to find the popultaions, which come to be 5360 million in 1993 and 8524 million in 2020. As you can see, it lines up with the graph as expontential growth.
3) It is growth (exponentially)
-Allison Moore
1: I chose Example 1; which examined predicting population growth.
2: You start out with two known population sizes with respect to their year. In the example they choose the years 1950 and 1960 to have a population of 2560 million and 3040 million respectively. You set your early year, 1950, as t=0 and then second year, 1960, to t=10. Since theorem 2 tells us that
y(t)=y(0)e^kt
we just plug in our known variables to get
2560e^10/k= 3040
we solve for k and find that the relative growth rate is approximately 1.7% per year. Since we now know k, we can change the time to estimate a population size in the future or in the past. For instance if we wanted to predict the population size in 2050 we set t=100 and come out with an equation looking like this:
P(100)=2560e^(0.017)(100)
which equates to be roughly 260389 million.
3. The example I came up with is of course a model of exponential growth.
Trent Layton
1. Example 1: Pop. Growth
2. In this example the population is found within different years by using the equation P(t)=Pe^k(t). The given information is:
1950: 2560 million people
1960: 3040 million people
Therefore, 1950 is t=0 and 1960 is t=10. K is found to be 1.7% by setting the equation equal to 3040 and pluging 10 in for t. After k is found the population of any year is easily found. Simply subtract 1950 from the year you are looking for and use that difference as your t. Then plug the t and the k into the equation to find the population.
3. This is an example of exponential growth.
Michael Lyons
1. Example 1 (Population Growth)
2.First you need to start out with two know population sizes and the corresponding year. In the example in the book they used 1950 which had a population of approximately 2560 million. They also used the year 1960 which had a population of approximately 3040 million. Use the second therom y(t)= y(0)e^kt by setting the year 1950 to t=0 and the later year 1960 to t=10. Then plug in the variables that are known and you get 2560=e^10/k=3040. After that is done solve for k and the rate ends up being 1.7% per year. Since we know k we can now predict population increases in the future or in the past. All you have to do is subtract the year 1950 from the specific year you want to see. Then plug in the variables and you will find the answer you seek.
3.Example 1 is an exponential growth.
-David Schwartz
1. (Example 1) Population growth
2. The way this is used is they selected two years to compare in this case it was 1950 and 1960. They gathered the population of that time for either one consecutively 2560 and 3040. They are comparing how the population grew from 1950 to 1960 meaning that the t for 1950 will be 0 because it is the starting. 1960 is ten years after 1950 so the t will be equal to 10. You then implement these numbers into the equation y(t)= y(0)e^kt first for 1950 and then for 1960. Then you will have this 2560=e^10/k=3040, finally you solve for K. The rate then ends up being 1.7% per year. Once we achieve k we now may predict population increases in the future or what the population was in the past. This is possible by subtracting the year you want from 1950. You finally plug in the variables and can determine the population growth from that time.
3. The 1950 to 1960 was exponential growth. If it would have been from 1960 to 1950 then it would be exponential decay.
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